Question

What is the refractive index of the Prism when the critical angle is 450?
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Answer 1

Solution :

⏭ Given:

• Critical angle of prism = 45°

⏭ To Find:

• Refractive index of prism

⏭ Formula:

✏ Relation between refractive index of medium and critical angle is given by...

$\bigstar \: \boxed{ \bf{ \pink{ \large{ \mu = \dfrac{1}{ \sin{c}}}}}}$

⏭ Terms indication:

• $\mu$ denotes refractive index of medium
• C denotes critical angle

⏭ Calculation:

$\twoheadrightarrow \sf \: \mu = \dfrac{1}{ \sin45 \degree} \\ \\ \red{ \dag \tt \: \sin45 \degree = \dfrac{1}{ \sqrt{2} } } \\ \\ \twoheadrightarrow \sf \: \mu = \dfrac{1}{ \frac{1}{ \sqrt{2} } } \\ \\ \twoheadrightarrow \: \boxed{ \sf{ \purple{ \large{ \mu = \sqrt{2} = 1.41}}}}$

⏭ Additional information:

• Refractive index is an unitless as well as dimensionless quantity.

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

• Critical angle of prism = 45

We know that , the angle of incidence in the denser medium for which the angle of refraction in rarer medium is 90° is called critical angle

$\large \mathtt{ \fbox{Refractive \: index </p><p> = \frac{1}{ \sin i_{c} } }}$

Thus ,

$\sf \mapsto Refractive \: index = \frac{1}{ \sin(45) } \\ \\ \sf \mapsto Refractive \: index = \frac{ \sqrt{2} }{1} \\ \\\sf \mapsto Refractive \: index = \sqrt{2} \\ \\\sf \mapsto Refractive \: index = 1.414$

Hence , the refractive index is 1.414

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